[英]Intersection of Two LineStrings Geopandas
假設我有以下字符串的GeoDataFrames,其中一個代表道路,其中一個代表等高線。
>>> import geopandas as gpd
>>> import geopandas.tools
>>> import shapely
>>> from shapely.geometry import *
>>>
>>> r1=LineString([(-1,2),(3,2.5)])
>>> r2=LineString([(-1,4),(3,0)])
>>> Roads=gpd.GeoDataFrame(['Main St','Spruce St'],geometry=[r1,r2], columns=['Name'])
>>> Roads
Name geometry
0 Main St LINESTRING (-1 2, 3 2.5)
1 Spruce St LINESTRING (-1 4, 3 0)
>>>
>>> c1=LineString(Point(1,2).buffer(.5).exterior)
>>> c2=LineString(Point(1,2).buffer(.75).exterior)
>>> c3=LineString(Point(1,2).buffer(.9).exterior)
>>> Contours=gpd.GeoDataFrame([100,90,80],geometry=[c1,c2,c3], columns=['Elevation'])
>>> Contours
Elevation geometry
0 100 LINESTRING (1.5 2, 1.497592363336099 1.9509914...
1 90 LINESTRING (1.75 2, 1.746388545004148 1.926487...
2 80 LINESTRING (1.9 2, 1.895666254004977 1.9117845...
>>>
如果我繪制這些,它們看起來像這樣:
有3條輪廓線和2條道路。 我想在每條道路的每個點找到高程。 基本上我想要交叉道路和輪廓(這應該給我12分)並保留兩個地理位置(道路名稱和高程)的屬性。
我可以通過使用兩個地理數據框的聯合的交集來生成12個點:
>>> Intersection=gpd.GeoDataFrame(geometry=list(Roads.unary_union.intersection(Contours.unary_union)))
>>> Intersection
geometry
0 POINT (0.1118644118110415 2.13898305147638)
1 POINT (0.2674451642029509 2.158430645525369)
2 POINT (0.3636038969321072 2.636396103067893)
3 POINT (0.4696699141100895 2.530330085889911)
4 POINT (0.5385205980649126 2.192315074758114)
5 POINT (0.6464466094067262 2.353553390593274)
6 POINT (1.353553390593274 1.646446609406726)
7 POINT (1.399321982208571 2.299915247776072)
8 POINT (1.530330085889911 1.46966991411009)
9 POINT (1.636396103067893 1.363603896932107)
10 POINT (1.670759586114587 2.333844948264324)
11 POINT (1.827239686607525 2.353404960825941)
>>>
但是,我現在如何獲得這12個點中的每個點的道路名稱和高程? 空間連接的行為與我的預期不同,只返回4個點(所有12個應該與行文件相交,因為它們是按照定義創建的)。
>>> gpd.tools.sjoin(Intersection, Roads)
geometry index_right Name
2 POINT (0.3636038969321072 2.636396103067893) 1 Spruce St
3 POINT (0.4696699141100895 2.530330085889911) 1 Spruce St
5 POINT (0.6464466094067262 2.353553390593274) 1 Spruce St
6 POINT (1.353553390593274 1.646446609406726) 1 Spruce St
>>>
有關如何做到這一點的任何建議?
編輯:問題似乎與交叉點的創建方式有關。 如果我以非常小的量緩沖道路和輪廓,則交叉點按預期工作。 見下文:
>>> RoadsBuff=gpd.GeoDataFrame(Roads, geometry=Roads.buffer(.000005))
>>> ContoursBuff=gpd.GeoDataFrame(Contours, geometry=Contours.buffer(.000005))
>>>
>>> Join1=gpd.tools.sjoin(Intersection, RoadsBuff).drop('index_right',1).sort_index()
>>> Join2=gpd.tools.sjoin(Join1, ContoursBuff).drop('index_right',1).sort_index()
>>>
>>> Join2
geometry Name Elevation
0 POLYGON ((1.636395933642091 1.363596995290097,... Spruce St 80
1 POLYGON ((1.530329916464109 1.469663012468079,... Spruce St 90
2 POLYGON ((1.353553221167472 1.646439707764716,... Spruce St 100
3 POLYGON ((0.5385239436706243 2.192310454047735... Main St 100
4 POLYGON ((0.2674491823047923 2.158426108877007... Main St 90
5 POLYGON ((0.1118688004427904 2.138978561144256... Main St 80
6 POLYGON ((0.6464467873602107 2.353546141571978... Spruce St 100
7 POLYGON ((0.4696700920635739 2.530322836868614... Spruce St 90
8 POLYGON ((0.3636040748855915 2.636388854046597... Spruce St 80
9 POLYGON ((1.399312865255344 2.299919147068011,... Main St 100
10 POLYGON ((1.670752113626148 2.333849053114361,... Main St 90
11 POLYGON ((1.827232214119086 2.353409065675979,... Main St 80
>>>
上面是所需的輸出,雖然我不確定為什么我必須緩沖線以使它們與從線的交叉點創建的點相交。
請注意,操作unary_union
和intersection
是在GeoDataFrame
內部的幾何上進行的,因此您將丟失存儲在其余列中的數據。 我認為在這種情況下,您必須通過訪問數據框中的每個幾何圖形來手動完成。 以下代碼:
import geopandas as gpd
from shapely.geometry import LineString, Point
r1=LineString([(-1,2),(3,2.5)])
r2=LineString([(-1,4),(3,0)])
roads=gpd.GeoDataFrame(['Main St','Spruce St'],geometry=[r1,r2], columns=['Name'])
c1=LineString(Point(1,2).buffer(.5).exterior)
c2=LineString(Point(1,2).buffer(.75).exterior)
c3=LineString(Point(1,2).buffer(.9).exterior)
contours=gpd.GeoDataFrame([100,90,80],geometry=[c1,c2,c3], columns=['Elevation'])
columns_data = []
geoms = []
for _, n, r in roads.itertuples():
for _, el, c in contours.itertuples():
intersect = r.intersection(c)
columns_data.append( (n,el) )
geoms.append(intersect)
all_intersection = gpd.GeoDataFrame(columns_data, geometry=geoms,
columns=['Name', 'Elevation'])
print all_intersection
生產:
Name Elevation geometry
0 Main St 100 (POINT (0.5385205980649125 2.192315074758114),...
1 Main St 90 (POINT (0.2674451642029509 2.158430645525369),...
2 Main St 80 (POINT (0.1118644118110415 2.13898305147638), ...
3 Spruce St 100 (POINT (0.6464466094067262 2.353553390593274),...
4 Spruce St 90 (POINT (0.4696699141100893 2.53033008588991), ...
5 Spruce St 80 (POINT (0.363603896932107 2.636396103067893), ...
請注意,每個幾何體都有兩個點,如果您想要逐點信息,可以稍后訪問這些點,或者您可以為每個點創建一個行,引入迭代點的for循環,如:
for p in intersect:
columns_data.append( (n,el) )
geoms.append(p)
但在這種情況下,你依賴於知道每個交叉點產生一個多幾何。
關於使用sjoin
函數的其他方法,我無法測試它,因為我正在使用的geopandas
版本不提供tools
模塊。 嘗試放入buffer(0.0)
以查看會發生什么。
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